Integrand size = 174, antiderivative size = 28 \[ \int \frac {e^{5 e^x x+5 x^2+e^{e^{e^{10 x}}} \left (5 e^x+5 x\right )} \left (e^{2+x} (-5-5 x)-10 e^2 x+e^{e^{e^{10 x}}} \left (-5 e^2-5 e^{2+x}+e^{e^{10 x}+10 x} \left (-50 e^{2+x}-50 e^2 x\right )\right )\right )}{9+6 e^{5 e^x x+5 x^2+e^{e^{e^{10 x}}} \left (5 e^x+5 x\right )}+e^{10 e^x x+10 x^2+2 e^{e^{e^{10 x}}} \left (5 e^x+5 x\right )}} \, dx=\frac {e^2}{3+e^{5 \left (e^{e^{e^{10 x}}}+x\right ) \left (e^x+x\right )}} \]
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Time = 8.97 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {6873, 6843, 32} \[ \int \frac {e^{5 e^x x+5 x^2+e^{e^{e^{10 x}}} \left (5 e^x+5 x\right )} \left (e^{2+x} (-5-5 x)-10 e^2 x+e^{e^{e^{10 x}}} \left (-5 e^2-5 e^{2+x}+e^{e^{10 x}+10 x} \left (-50 e^{2+x}-50 e^2 x\right )\right )\right )}{9+6 e^{5 e^x x+5 x^2+e^{e^{e^{10 x}}} \left (5 e^x+5 x\right )}+e^{10 e^x x+10 x^2+2 e^{e^{e^{10 x}}} \left (5 e^x+5 x\right )}} \, dx=-\frac {e^2}{3 \left (3 e^{-5 \left (x+e^{e^{e^{10 x}}}\right ) \left (x+e^x\right )}+1\right )} \]
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Rule 32
Rule 6843
Rule 6873
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{5 \left (e^{e^{e^{10 x}}}+x\right ) \left (e^x+x\right )} \left (e^{2+x} (-5-5 x)-10 e^2 x+e^{e^{e^{10 x}}} \left (-5 e^2-5 e^{2+x}+e^{e^{10 x}+10 x} \left (-50 e^{2+x}-50 e^2 x\right )\right )\right )}{\left (3+e^{5 \left (e^{e^{e^{10 x}}}+x\right ) \left (e^x+x\right )}\right )^2} \, dx \\ & = \frac {1}{3} e^2 \text {Subst}\left (\int \frac {1}{(1+x)^2} \, dx,x,3 e^{-5 \left (e^{e^{e^{10 x}}}+x\right ) \left (e^x+x\right )}\right ) \\ & = -\frac {e^2}{3 \left (1+3 e^{-5 \left (e^{e^{e^{10 x}}}+x\right ) \left (e^x+x\right )}\right )} \\ \end{align*}
Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{5 e^x x+5 x^2+e^{e^{e^{10 x}}} \left (5 e^x+5 x\right )} \left (e^{2+x} (-5-5 x)-10 e^2 x+e^{e^{e^{10 x}}} \left (-5 e^2-5 e^{2+x}+e^{e^{10 x}+10 x} \left (-50 e^{2+x}-50 e^2 x\right )\right )\right )}{9+6 e^{5 e^x x+5 x^2+e^{e^{e^{10 x}}} \left (5 e^x+5 x\right )}+e^{10 e^x x+10 x^2+2 e^{e^{e^{10 x}}} \left (5 e^x+5 x\right )}} \, dx=\frac {e^2}{3+e^{5 \left (e^{e^{e^{10 x}}}+x\right ) \left (e^x+x\right )}} \]
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Time = 97.84 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82
method | result | size |
risch | \(\frac {{\mathrm e}^{2}}{{\mathrm e}^{5 \left ({\mathrm e}^{{\mathrm e}^{{\mathrm e}^{10 x}}}+x \right ) \left ({\mathrm e}^{x}+x \right )}+3}\) | \(23\) |
parallelrisch | \(\frac {{\mathrm e}^{2}}{{\mathrm e}^{\left (5 \,{\mathrm e}^{x}+5 x \right ) {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{10 x}}}+5 \,{\mathrm e}^{x} x +5 x^{2}}+3}\) | \(37\) |
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Timed out. \[ \int \frac {e^{5 e^x x+5 x^2+e^{e^{e^{10 x}}} \left (5 e^x+5 x\right )} \left (e^{2+x} (-5-5 x)-10 e^2 x+e^{e^{e^{10 x}}} \left (-5 e^2-5 e^{2+x}+e^{e^{10 x}+10 x} \left (-50 e^{2+x}-50 e^2 x\right )\right )\right )}{9+6 e^{5 e^x x+5 x^2+e^{e^{e^{10 x}}} \left (5 e^x+5 x\right )}+e^{10 e^x x+10 x^2+2 e^{e^{e^{10 x}}} \left (5 e^x+5 x\right )}} \, dx=\text {Timed out} \]
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Time = 0.45 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {e^{5 e^x x+5 x^2+e^{e^{e^{10 x}}} \left (5 e^x+5 x\right )} \left (e^{2+x} (-5-5 x)-10 e^2 x+e^{e^{e^{10 x}}} \left (-5 e^2-5 e^{2+x}+e^{e^{10 x}+10 x} \left (-50 e^{2+x}-50 e^2 x\right )\right )\right )}{9+6 e^{5 e^x x+5 x^2+e^{e^{e^{10 x}}} \left (5 e^x+5 x\right )}+e^{10 e^x x+10 x^2+2 e^{e^{e^{10 x}}} \left (5 e^x+5 x\right )}} \, dx=\frac {e^{2}}{e^{5 x^{2} + 5 x e^{x} + \left (5 x + 5 e^{x}\right ) e^{e^{e^{10 x}}}} + 3} \]
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Time = 0.45 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {e^{5 e^x x+5 x^2+e^{e^{e^{10 x}}} \left (5 e^x+5 x\right )} \left (e^{2+x} (-5-5 x)-10 e^2 x+e^{e^{e^{10 x}}} \left (-5 e^2-5 e^{2+x}+e^{e^{10 x}+10 x} \left (-50 e^{2+x}-50 e^2 x\right )\right )\right )}{9+6 e^{5 e^x x+5 x^2+e^{e^{e^{10 x}}} \left (5 e^x+5 x\right )}+e^{10 e^x x+10 x^2+2 e^{e^{e^{10 x}}} \left (5 e^x+5 x\right )}} \, dx=\frac {e^{2}}{e^{\left (5 \, x^{2} + 5 \, x e^{x} + 5 \, x e^{\left (e^{\left (e^{\left (10 \, x\right )}\right )}\right )} + 5 \, e^{\left (x + e^{\left (e^{\left (10 \, x\right )}\right )}\right )}\right )} + 3} \]
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Timed out. \[ \int \frac {e^{5 e^x x+5 x^2+e^{e^{e^{10 x}}} \left (5 e^x+5 x\right )} \left (e^{2+x} (-5-5 x)-10 e^2 x+e^{e^{e^{10 x}}} \left (-5 e^2-5 e^{2+x}+e^{e^{10 x}+10 x} \left (-50 e^{2+x}-50 e^2 x\right )\right )\right )}{9+6 e^{5 e^x x+5 x^2+e^{e^{e^{10 x}}} \left (5 e^x+5 x\right )}+e^{10 e^x x+10 x^2+2 e^{e^{e^{10 x}}} \left (5 e^x+5 x\right )}} \, dx=\text {Timed out} \]
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Time = 8.66 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \[ \int \frac {e^{5 e^x x+5 x^2+e^{e^{e^{10 x}}} \left (5 e^x+5 x\right )} \left (e^{2+x} (-5-5 x)-10 e^2 x+e^{e^{e^{10 x}}} \left (-5 e^2-5 e^{2+x}+e^{e^{10 x}+10 x} \left (-50 e^{2+x}-50 e^2 x\right )\right )\right )}{9+6 e^{5 e^x x+5 x^2+e^{e^{e^{10 x}}} \left (5 e^x+5 x\right )}+e^{10 e^x x+10 x^2+2 e^{e^{e^{10 x}}} \left (5 e^x+5 x\right )}} \, dx=\frac {{\mathrm {e}}^2}{{\mathrm {e}}^{5\,x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{5\,x\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^{10\,x}}}}\,{\mathrm {e}}^{5\,x^2}\,{\mathrm {e}}^{5\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^{10\,x}}}\,{\mathrm {e}}^x}+3} \]
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